Wikipedia:Reference desk/Archives/Mathematics/2013 February 12

= February 12 =

Cosets and normal subgroup equality
Hello. I recently came across an exercise, and would be very grateful if someone could help.

Given a normal subgroup H in G, with g in G, if $$gH = H$$, then is $$g \in H$$?

Neuroxic (talk) 11:31, 12 February 2013 (UTC)
 * Any subgroup H contains the identity. Think what happens with that. Dmcq (talk) 12:11, 12 February 2013 (UTC)
 * That question is a special case of the following question, which may actually be easier (intuitively) to solve: if $$gH = L$$, then is $$g \in L$$? Basically you're just asking "Is $$g \in gH$$ for some subgroup $$H$$?" and it is a very well known fact that the answer is yes, and the proof is also very well known and straightforward: $$1 \in H \implies g\cdot 1 \in gH \implies g \in gH$$ --AnalysisAlgebra (talk) 19:21, 12 February 2013 (UTC)
 * There's no need for $$H$$ to be normal, by the way. --AnalysisAlgebra (talk) 19:23, 12 February 2013 (UTC)

Estimating parameters of a distribution from censored data
Suppose you hypothesize that your data are drawn from, say, a lognormal distribution or a gamma distribution. Your data are right-censored at the value x*. That is, you have exact data whenever x≤x*, but whenever x>x* all you know is that x>x*.

(1) How do you estimate the parameters of the hypothesized distribution? I know it would be by maximum likelihood, but how do you handle the censored data?

(2) Is there a command in, say, SAS that will do this?

(3) How do you test the hypothesis that the data did in fact come from that distribution? Duoduoduo (talk) 18:37, 12 February 2013 (UTC)
 * (1) If your data consists of uncensored points $$x_1,\ldots,x_m$$ and n censored points, and your pdf with parameters $$\theta$$ is $$f(x|\theta)$$, then the likelihood of $$\theta$$ is $$\left(\int_{x^*}^{\infty}f(t|\theta)\ dt\right)^n\prod_{i=1}^mf(x_i|\theta)$$. There's ostensibly an inconsistency in that you use a density for the contribution to likelihood of some points and a probability for others, but since likelihood functions are only meaningful up to a constant factor it doesn't matter. -- Meni Rosenfeld (talk) 19:49, 12 February 2013 (UTC)


 * Thanks, Meni! Duoduoduo (talk) 14:19, 14 February 2013 (UTC)